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Laplace's Method Integration

  1. Apr 19, 2014 #1

    wel

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    Gold Member

    Consider the integral
    \begin{equation}
    I_n(x)=\int^{2}_{1} (log_{e}t) e^{-x(t-1)^{n}}dt
    \end{equation}
    Use Laplace's Method to show that
    \begin{equation}
    I_n(x) \sim \frac{1}{nx^\frac{2}{n}} \int^{\infty}_{0} \tau^{\frac{2-n}{n}} e^{-\tau} d\tau \end{equation}
    as [itex]x\rightarrow\infty[/itex].
    where [itex]0<n\leq2[/itex]. Hence find the leading order behaviour of [itex]I_{1}(x)[/itex]. and [itex]I_{2}(x)[/itex] as [itex]x\rightarrow \infty[/itex].

    =>
    Its really difficult question for me.

    Here,

    [itex]g(t) = -(t-1)^{n}[/itex] has the maximum at [itex]t=0[/itex]

    but [itex]h(t)= log_{e}t[/itex] at [itex]t=0[/itex]
    [itex]h(0)=0[/itex].
    so I can not go any further. PLEASE HELP ME.
     
  2. jcsd
  3. Apr 20, 2014 #2
    I believe you meant [itex]g[/itex] reaches its maximum at [itex]t = 1[/itex].
    Indeed, [itex]h(1) = 0[/itex], but that shouldn't be a problem, just use higher order terms in the Taylor expansion of the logarithm.
     
  4. Apr 20, 2014 #3

    wel

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    Gold Member

    at [itex] t=1 [/itex], [itex] g(t) =0 [/itex], how can i say it is maximum?

    what is the Taylor expansion of the [itex] log_{e} t[/itex]?
     
  5. Apr 21, 2014 #4
    [itex] g [/itex] is decreasing.

    Do you remember what a Taylor series is? That's basic if you want to calculate asymptotic expansions of integrals.
     
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